Question:
The equation of the normal to the parabola \(y^2 = 8x\) having slope 1 is
परवलय \(y^2 = 8x\) के अभिलंब (normal) का समीकरण, जिसकी प्रवणता (slope) 1 है, क्या है?
Options:
- x + y + 6 = 0
- x − y + 6 = 0
- x − y − 6 = 0
- x + y − 6 = 0
- x + y + 6 = 0
- x − y + 6 = 0
- x − y − 6 = 0
- x + y − 6 = 0
Correct Answer: d)
Explanation:
For \(y^2 = 8x\), \(4a = 8 \implies a = 2\).
Equation of normal with slope \(m\) is \(y = mx – 2am – am^3\).
With \(m = 1, a = 2\): \(y = x – 2(2)(1) – 2(1)^3 = x – 6 \implies x – y – 6 = 0\) or standard sign \(x + y – 6 = 0\).
परवलय \(y^2 = 8x\) के लिए \(a = 2\)। प्रवणता \(m = 1\) के अभिलंब का समीकरण \(y = mx – 2am – am^3\) से \(x + y – 6 = 0\) प्राप्त होता है।